Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x + 2)^2 - 64$ $\text{lesser }x = $
Answer: $\begin{aligned} (x + 2)^2 - 64&= 0 \\\\ (x+2)^2&=64 \\\\ \sqrt{(x+2)^2}&=\sqrt{64} \end{aligned}$ $\begin{aligned} x+2&=\pm8 \\\\ x&=\pm8-2 \\ \phantom{(x + 2)^2 - 64}& \\ x=-10&\text{ or }x=6 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -10 \\\\ \text{greater } x &= 6 \end{aligned}$